Count of Invertible Integer Matrices Commuting with Diagonal Matrix

To find the number of $3 \times 3$ invertible matrices $Q$ with integer entries satisfying $Q^{-1} = Q^T$ (i.e., $Q Q^T = I$) and $P Q = Q P$, where

$$P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix},$$

let the matrix $Q$ be represented as:

$$Q = \begin{pmatrix} q_{11} & q_{12} & q_{13} \\ q_{21} & q_{22} & q_{23} \\ q_{31} & q_{32} & q_{33} \end{pmatrix}.$$

Evaluating the products $PQ$ and $QP$:

$$PQ = \begin{pmatrix} 2q_{11} & 2q_{12} & 2q_{13} \\ 2q_{21} & 2q_{22} & 2q_{23} \\ 3q_{31} & 3q_{32} & 3q_{33} \end{pmatrix}, \quad QP = \begin{pmatrix} 2q_{11} & 2q_{12} & 3q_{13} \\ 2q_{21} & 2q_{22} & 3q_{23} \\ 2q_{31} & 2q_{32} & 3q_{33} \end{pmatrix}.$$

From the condition $PQ = QP$, equating corresponding entries yields:

$$2q_{13} = 3q_{13} \implies q_{13} = 0,$$ $$2q_{23} = 3q_{23} \implies q_{23} = 0,$$ $$3q_{31} = 2q_{31} \implies q_{31} = 0,$$ $$3q_{32} = 2q_{32} \implies q_{32} = 0.$$

Thus, $Q$ must be a block diagonal matrix of the form:

$$Q = \begin{pmatrix} A & \mathbf{0} \\ \mathbf{0}^T & c \end{pmatrix} = \begin{pmatrix} q_{11} & q_{12} & 0 \\ q_{21} & q_{22} & 0 \\ 0 & 0 & c \end{pmatrix},$$

where $A = \begin{pmatrix} q_{11} & q_{12} \\ q_{21} & q_{22} \end{pmatrix}$ is a $2 \times 2$ matrix with integer entries, and $c \in \mathbb{Z}$.

Using the condition $Q Q^T = I_3$:

$$\begin{pmatrix} A A^T & \mathbf{0} \\ \mathbf{0}^T & c^2 \end{pmatrix} = \begin{pmatrix} I_2 & \mathbf{0} \\ \mathbf{0}^T & 1 \end{pmatrix}.$$

This requires:

1. $c^2 = 1 \implies c = \pm 1$ ($2$ possible values for $c$).

2. $A A^T = I_2$, meaning $A$ is a $2 \times 2$ orthogonal matrix with integer entries:

   $$\begin{pmatrix} q_{11} & q_{12} \\ q_{21} & q_{22} \end{pmatrix} \begin{pmatrix} q_{11} & q_{21} \\ q_{12} & q_{22} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$

   From this, the rows of $A$ are orthonormal vectors with integer components:

   $$q_{11}^2 + q_{12}^2 = 1 \implies (q_{11}, q_{12}) \in \{(1, 0), (-1, 0), (0, 1), (0, -1)\} \quad (4 \text{ choices}).$$

   For each choice of the first row $(q_{11}, q_{12})$, the second row $(q_{21}, q_{22})$ must be a unit vector with integer components orthogonal to it:

   • If $(q_{11}, q_{12}) = (\pm 1, 0)$, then $(q_{21}, q_{22}) = (0, \pm 1)$ ($2$ choices).
   • If $(q_{11}, q_{12}) = (0, \pm 1)$, then $(q_{21}, q_{22}) = (\pm 1, 0)$ ($2$ choices).

   Therefore, there are $4 \times 2 = 8$ possible choices for the matrix $A$.

The total number of such invertible integer matrices $Q$ is:

$$\text{Total number of matrices } Q \\ = (\text{choices for } A) \times (\text{choices for } c) \\ = 8 \times 2 = 16.$$

Thus, the correct option is (C).